I'm studying for a qualifying exam (tomorrow) and I was hoping someone could show me how to finish solving this problem.
Let \begin{align} a_n = 1 - \frac{1}{n^2}, && f(z) = \prod_{n=1}^\infty \frac{a_n - z}{1 - a_n z}. \end{align} Show $f(z)$ is analytic on the unit disk $\left\{ z \colon \left| z \right| < 1 \right\}$ but cannot be analytically extended to any larger disk $\left\{ z \colon \left| z \right| < r \right\}$ for any $r > 1$.
I was able to prove $f(z)$ is analytic on the unit disk by writing $f(z) = \prod 1 + A_n$, where $$ A_n = \frac{-\left( 1 + z \right)}{n^2 + \left( 1 - n^2 \right) z} $$ from which it follows fairly easily that $\sum \left| A_n \right|$ converges uniformly on compact sets.
I was hoping that the second part would follow by letting $z = x$ tend to $1^-$. But one sees that doing so makes $\log f(x) = \sum 1 + A_n$ tend to $-\infty$, which means $f(x)$ tends to $0$. At least a priori there is no reason why an extension can't exist that has a zero at $x = 1$. I briefly considered letting $z$ tend to other values on the unit circle, but didn't see anything promising.
Hopefully someone can help me out tonight! Thanks a lot!
The second part.
Suppose that $f(z)$ can be analytically extended to some larger disk $\left\{ z \colon \left| z \right| < r \right\}$ for $r>1$ and let $F(z)$ be the extention. Then $(a_n)$ is the sequence of zeros of $F(z)$, by the definition of $f(z)$, which accumulates at $z=1$. Therefore $z=1$ must be an essential singurality of $F(z)$. This is a contradiction.